两异面直线L1:x=y-=乙 L2:x2=2=22间的距离 d=rr/a1×a2 MM2 1 1Xa2)M,M 2 2 2-x1y2-y1Z2 a 2 2
: , 1 1 1 1 1 1 1 p z z n y y m x x L − = − = − 两异面直线 : : 2 2 2 2 2 2 2 间的距离 p z z n y y m x x L − = − = − → = Pr 1 2 M1M2 d j 1 2 1 2 1 2 ( ) = → M M . 2 2 2 1 1 1 2 1 2 1 2 1 2 2 2 1 1 1 m n p m n p i j k x x y y z m n p m n p − − − =
(5)平面及直线间的位置关系 平面与平面: 丌1:A1x+B1y+C1z+D1=0, 丌2:A2x+B2y+C2x+D2=0, cosO=吃 2 A12+B1B2+C1C2 A12+B 2 2 2 2 丌1⊥x2分A1A2+B1B2+C1C2=0; B C ∥2x2分→ B2 K
(5) 平面及直线间的位置关系 平面与平面: : 0, 1 A1x + B1 y + C1 z + D1 = : 0, 2 A2 x + B2 y + C2 z + D2 = 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 | | cos A B C A B C A A B B C C + + + + + + = = 1 ⊥ 2 0; A1A2 + B1B2 + C1C2 = . 2 1 2 1 2 1 C C B B A A 1 // 2 = =
直线与直线: x-xi y-y1 Z-Z x-x2y-y2_-2 m1m2+n12+n1P2 cos 19-2 2 2 2 22 2 n1+n1 十 十 L1⊥L2<→m1m2+n12+p1P2=0, L1∥L2<→ n p 2 1y2-y1z2 L1与L2共面∈→m n P1=0 K
直线与直线: : , 1 1 1 1 1 1 1 p z z n y y m x x L − = − = − : , 2 2 2 2 2 2 2 p z z n y y m x x L − = − = − 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 2 1 2 | | cos( , ) m n p m n p m m n n p p L L + + + + + + ^ = L1 ⊥ L2 0, m1m2 + n1n2 + p1 p2 = , 2 1 2 1 2 1 p p n n m m L1 // L2 = = L1与L2共面 0 2 2 2 1 1 1 2 1 2 1 2 1 = − − − m n p m n p x x y y z z
平面与直线: L:0=y-yo z-10, E: Ax+By+Cz+D=0, Am+ Bn+Cp Sin p √A2+B2+C2·√m2+n2+p AB C L⊥兀仝→ L∥冷Am+Bn+Cp=0 Lc丌分Am+Bn+p=0,且4x+B+C+D=0 Co+ mt 0 已知π与L,求交点:令{y=+mt, zo t pt 0 代入Ax+B+Cz+D=0得,从而可得交点.图四圆
平面与直线: : , 0 0 0 p z z n y y m x x L − = − = − : Ax + By +Cz + D = 0, 2 2 2 2 2 2 | | sin A B C m n p Am Bn Cp + + + + + + = L ⊥ . p C n B m A = = L// Am+ Bn+Cp = 0. L Am+ Bn+Cp = 0,且Ax0 + By0 +Cz0 + D = 0 已知与L, 求交点: , 0 0 0 = + = + = + z z pt y y nt x x mt 令 代入Ax + By +Cz + D = 0得t,从而可得交点